1

I have a question where I have to prove that $x + y$ is not a multiple of $9$, assuming that $x$ is a multiple of $9$ and $y$ is not a multiple of $3$. I know I can just substitute values to prove this, but I'm trying to do it algebraically. I was wondering if anyone can help me out here.

Thanks

Masacroso
  • 30,417
Broadsword93
  • 567
  • 6
  • 21
  • 5
    If $x+y$ is divisible by $9$ and $x$ is divisible by $9$ then $y=(x+y)-x$ is divisible by $9$. – Thomas Andrews Jul 28 '17 at 02:19
  • First you need to write algebraically what "$x$ is a multiple of $K$" means when $K\ne 0$, which is that $x/K \in \mathbb Z.$ So if $x$ and $x+y$ are multiples of $9$ then $x/9\in \mathbb Z$ and $(x+y)/9 \in \mathbb Z ,$ implying $y/9=(x+y)/9-x/9$ is the difference of two integers, so $y/9\in \mathbb Z,$........ so $y/3=3(y/9)$ is $3$ times an integer so $y/3 \in \mathbb Z$ so $y$ is divisible by $3$. – DanielWainfleet Jul 28 '17 at 07:31

1 Answers1

1

Proof by contradiction.

Assume $x+y $ and $x $ are both multiples of $9$ .

Then $x =9k $ and $x+y=9j$ for integer $k,j $

So $x+y=9j $

$9k + y =9j$

$y= 9j-9k=3 (3j-3k) $.

So $y$ is a multiple of $3$ .

Which contradicts our assumption.

fleablood
  • 124,253